Janek191:
f(x) = x
2 + 4 x − 2 < − 3; 2 >
Mamy
a = 1, b = 4 , c = −2
więc
| | − b | | − 4 | |
p = |
| = |
| = − 2 ∊ < − 3; 2 > |
| | 2a | | 2 | |
a = 1 > 0 ∧ p ∊ < − 3; 2 > więc
y
min = f(p) = f(− 2) = ( −2)
2 + 4*(−2) − 2 = 4 − 8 − 2 = − 6
oraz
y
max = f( 2) = 2
2 + 4*2 − 2 = 4 + 8 − 2 = 10
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